Definite Integration Question 374
Question: $ \int_0^{\pi /2}{\frac{\cos x}{1+\cos x+\sin x}}dx= $
[Roorkee 1989]
Options:
A) $ \frac{\pi }{4}+\frac{1}{2}\log 2 $
B) $ \frac{\pi }{4}+\log 2 $
C) $ \frac{\pi }{4}-\frac{1}{2}\log 2 $
D) $ \frac{\pi }{4}-\log 2 $
Show Answer
Answer:
Correct Answer: C
Solution:
$ \int_0^{\pi /2}{\frac{\cos x}{1+\cos x+\sin x}}dx $
$ =\int_0^{\pi /2}{\frac{{{\cos }^{2}}(x/2)-{{\sin }^{2}}(x/2)}{2{{\cos }^{2}}(x/2)+2\sin (x/2)\cos (x/2)}}dx $
$ =\frac{1}{2}\int_0^{\pi /2}{\frac{1-{{\tan }^{2}}(x/2)}{1+\tan (x/2)}}dx=\frac{1}{2}\int_0^{\pi /2}{[ 1-\tan ( \frac{x}{2} ) ]}dx $
$ \frac{\pi }{4}+\log \frac{1}{\sqrt{2}}=\frac{\pi }{4}-\frac{1}{2}\log 2 $ .
BETA
